Triangle

There are two ways to do this...
One of the easiest way is,
since, LB = 2LC
so, b² = c (c + a)
as c = n so,
or, b² = n² + an ---- (i)
Again, by angle bisector theorem,
c/b = m/n
but, c = n
=> n² = b x m ---- (ii)
putting, the value of n² from eqn. (ii) in (i) we get,
b² = bm + an
Again, a = (m + n)
so, b² = bm + n(m + n)
or, b² = bm + mn + n²
=> b² - n² = m (b + n)
=> (b + n)(b - n) = m (b + n)
=> b - n = m
=> b = (m + n)
what???
a = b
so, LA = LB
Let, LC = x so,
LB = LA = 2x
so, 2x + 2x + x = 180°
=> 5x = 180°
=> x = 36°
so, LA = 2x = 2(36°) = 72° proved..

Algeometry

"since angle B = 2 angle C so, b²= c(a+c)"

show it ; (ISI entrance 2007 objective)

Let ABC be the triangle in which LA = 2LB
Now, we draw AD from A meeting BC at D such that it bisects A
let, LB = x (say) ,
so, LA = 2x
=> LBAD = LCAD = x (by our construction)
Also, by angle-bisector theorem
AB/AC = BD/CD
clearly, AB = c , AC = b and BC = a
so, c/b = BD/CD
=> (b + c)/b = a/CD ---- (i)
Again, Triangle ADC ~ Triangle BAC clearly,
so, AD/AB = CD/AC = AC/BC
or, AD/c = CD/b = b/a
=> CD = b²/a ---- (ii)
putting the value of CD in (i) we get
(b + c)/b = a/(b²/a)
=> a²/b² = (b + c)/b
=> a² = b (b + c) proved... :D
What is ISI by the way??

There is one more way to work this question out..

>> Its by pure geometry.

Dare to be different, go a little crazy

i like it!

I like that u like it, but it's copied !

@Subhodip - What r u talking bout???

If u r talking bout the script... then its a self designed one of mine...
i've written some html codes for it... just for fun... !!
i've used a simple programming in JS....!!
onclick = 'alert(".....")'

:D